Consider a simple linear regression model with response $y$, an intercept and an explanatory variable $x$,i.e. $$y_i=\beta_0+\beta_1x_i+\epsilon_i, \;\;\;\;\;\;\;\; i=1,...,n.$$ Assume further that the residual variance proportional to the square of $x_i$: $V(\epsilon_i)=\sigma^2x^2_i$. Consider now the following data transformation: $$\tilde{y}_i:=\frac{y_i}{x_i} \text{ and } \tilde{x_i}:=\frac{1}{x_i}, \;\;\;\;\;\;\;\; i=1,...,n.$$ (1) Is this a variance-stabilizing transformation?

(2) How are the parameters from the models with original or transformed data related to each other?

Am i doing it correctly? i am not sure in (1) that : it is a variance-stabilizing transformation or not.

Answer (1) : Dividing $y_i$ by $x_i$ ,the model becomes $$ \tilde{y}_i=\beta_1+\beta_0\tilde{x}_i+\epsilon_i\tilde{x}_i $$ $$ \tilde{y}_i=\beta_1+\beta_0\tilde{x}_i+u_i $$ here $u_i:=\epsilon_i\tilde x_i$ $\to$ (residual term). $$ \operatorname{Var}(u_i)=\operatorname{Var}(\epsilon_i\tilde x_i) =\tilde x_i^2\operatorname{Var}(\epsilon_i) =\frac 1{x_i^2}.{\sigma^2x_i^2} =\sigma^2$$,for $i= 1,.....,n$ So, the variance of residual becomes constant for $i= 1,.....,n$ by using the above transformation. so, this is a variance-stabilizing transformation.

Answer (2): The original model $$y_i=\beta_0+\beta_1x_i+\epsilon_i$$After transformation:$$ \tilde{y}_i=\beta_1+\beta_0\tilde{x}_i+\epsilon_i\tilde{x}_i $$ (a) Intercept $(\beta_0)$

Now, multiplying orginal model by $\frac 1 {x_i} $ we get:$$ \frac {y_i}{x_i}=\frac {\beta_0}{x_i}+\beta_1+\frac{\epsilon_i} {x_i}$$ Comparing this with the transformed model,we see, $\beta_0$ in original model corresponds to $\frac {\beta_0}{x_i}$ in the transformed model.

(b)Slope $(\beta_1)$

The slope remains the same in both models, i.e.,$(\beta_1)$ in the original model is the same as $(\beta_1)$ in the transformed model.

So, the parameters are related through a scaling factor of $\frac {1}{x_i}$ for the intercept.


1 Answer 1


The first part is fine, but the second part is not quite correct. You actually wrote the correct model after transformation: $$\tilde y_i = \beta_1 + \beta_0 \tilde x_i + \epsilon_i \tilde x_i$$ with the error term $\tilde \epsilon_i = \epsilon_i \tilde x_i$ now being homoscedastic.

But the key insight here that is not stated is that the roles of slope and intercept have interchanged: the slope parameter of the transformed model is the intercept of the original, and the intercept parameter of the transformed model is the slope of the original. That is to say, letting $$\tilde \beta_0 = \beta_1, \quad \tilde \beta_1 = \beta_0,$$ we can now write the transformed model as $$\tilde y_1 = \tilde \beta_0 + \tilde \beta_1 \tilde x_i + \tilde \epsilon_i.$$

That is what the problem is trying to get you to see.

  • $\begingroup$ Thank you, i got the point, can i know more that; if we use weighted least squares with weights $w_i = \frac {1}{x^2_i}$. Is this equivalent to the above considered transformation? @heropup $\endgroup$
    – User1
    Commented Dec 8, 2023 at 13:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .